Problem: The equation of a circle $C$ is $x^2+y^2+4y-12 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2) + (y^2+4y) = 12$ $(x^2) + (y^2+4y+4) = 12 + 0 + 4$ $x^2 + (y+2)^{2} = 16 = 4^2$ Thus, $(h, k) = (0, -2)$ and $r = 4$.